# When does $\pi_1(\Sigma)$ inject into $\pi_1(S^3 \setminus \Sigma)$?

Here's a fun fact from knot theory:

$\quad$ If $\, \Sigma$ is a minimal-genus Seifert surface for a knot $K$, then $i_*:\pi_1(S^3 \setminus \Sigma) \to \pi_1(S^3 \setminus K)$ is injective, where $i: S^3 \setminus \Sigma \to S^3 \setminus K$ is the inclusion.

For a proof, see the "Lemma" in my second answer to this other question. It proves a slightly stronger result: If $\, \Sigma$ is a connected, compact, orientable surface in $S^3$ and $\pi_1(\Sigma^\pm)\to \pi_1(S^3 \setminus \Sigma)$ is injective, then $\pi_1(S^3 \setminus \Sigma) \to \pi_1(S^3 \setminus \partial \Sigma)$ is injective. With the goal of extending the first fact to links with multiple components, I ask:

When does $\pi_1(\Sigma^\pm)$ inject into $\pi_1(S^3 \setminus \Sigma)$? In particular, under what conditions does this hold for a minimal-genus Seifert surface of a link?

Being a minimal-genus Seifert surface is not enough, as is demonstrated by considering the two-component unlink bounding a standard annulus. There are some sufficient conditions:

Examples:

1. $\Sigma$ is a minimal Seifert surface for a knot, i.e. the first fact.

2. $\Sigma$ is a minimal Seifert surface for a two- or three-component link $L$ in which each component has nonzero linking number with some other component. If $\pi_1(\Sigma^+) \to \pi_1(S^3 \setminus \Sigma)$ has nontrivial kernel, the loop theorem gives us a nontrivial curve in $\Sigma^+$ bounding an embedded disk in $S^3 \setminus \Sigma$. Cutting $\Sigma^+$ along the curve and capping with parallel copies of the disk produces an embedded compact orientable surface $\Sigma'$. If $\Sigma'$ is connected, it is a Seifert surface for $L$ with genus less than that of $\Sigma$, a contradiction. If $\Sigma'$ is disconnected and contains a closed component (which will not be $S^2$ by construction), we can remove it to lower genus and get another contradiction. Finally, if $\Sigma'$ is disconnected and contains no closed components, then one of its connected components has only one boundary component and is therefore a Seifert surface for one of the link components, say $L_1$. But since there is some other link component, say $L_2$, whose linking number with $L_1$ is nonzero, $L_2$ must intersect the surface component bounded by $L_1$. This contradicts the fact that $\Sigma'$ is embedded. We conclude that $\pi_1(\Sigma^\pm)\to \pi_1(S^3 \setminus \Sigma)$ is injective.

If $\Sigma$ is a minimal Seifert surface of a link $L = \partial \Sigma$ contained in $S^3$, then you can find a taut foliation in the complement of $L$ having $\Sigma$ has leaf (this is a theorem proved by Thurston during the eighties). On the other hand, if $\Sigma$ is a leaf of a taut foliation of a three-manifold $Y$ (in this case the complement of the link), then the inclusion $i : \Sigma \to Y$ induces an injection on the fundamental groups, and we are done.
In general, the injectivity condition you are looking for is equivalent to $\textit{incompressibility}$. There is a lot of standard three-manifold theory developed around this condition, see for example this book https://books.google.it/books/about/Algorithmic_and_Computer_Methods_for_Thr.html?id=bjcZAQAAIAAJ&hl=it
• I should note that this is a good answer to the stated question about the injectivity of the map $\pi_1(\Sigma)\to \pi_1(S^3 \setminus \Sigma)$ because $\pi_1(\Sigma) \to \pi_1(S^3 \setminus \partial \Sigma)$ factors through $\pi_1(\Sigma) \to \pi_1(S^3 \setminus \Sigma)$. – Kyle Jun 4 '15 at 13:18
• Oh yes, you where asking for the inclusion $\pi_1( \Sigma) \to \pi_1(S^3 \smallsetminus \Sigma)$. Sure your observation is important! – Antonio Alfieri Jun 4 '15 at 23:01